OFFSET
1,2
COMMENTS
Another way to describe this: starting with the binary representation and a counter set at one, count the 0's from right to left. Write a term equal to the counter for each "1" encountered.
A101211 is a similar sequence, with A005811 elements per row which maps natural numbers to compositions (ordered partitions).
There are two ways to consider this as a table: taking each partition as a row, or taking the partitions generated by 2^(n-1) through 2^n-1 as a row.
Taking the n-th row as multiple partitions, it consists of those partitions with the first hook size (largest part plus number of parts minus 1) equal to n. The number of integers in this n-th row is A001792(n-1), and the row sum is A049611.
Heinz numbers of the rows are A005940. - Gus Wiseman, Jan 17 2023
LINKS
Alois P. Heinz, Rows n = 1..12, flattened
FORMULA
Partition 2n is partition n with every part size increased by 1; partition 2n+1 is partition n with an additional part of size 1.
EXAMPLE
Row 4:
1000 [4]
1001 [3,1]
1010 [3,2]
1011 [2,1,1]
1100 [3,3]
1101 [2,2,1]
1110 [2,2,2]
1111 [1,1,1,1]
MAPLE
b:= proc(n) local c, l, m; l:=[][]; m:= n; c:=1;
while m>0 do if irem(m, 2, 'm')=0 then c:= c+1
else l:= c, l fi
od; l
end:
T:= n-> seq(b(i), i=2^(n-1)..2^n-1):
seq(T(n), n=1..7); # Alois P. Heinz, Sep 25 2015
MATHEMATICA
f[k_] := (bits = IntegerDigits[k, 2]; zerosCount = Reverse[ Accumulate[ 1-Reverse[bits] ] ] + 1; Select[ Transpose[ {bits, zerosCount} ], First[#] == 1 & ][[All, 2]]); row[n_] := Table[ f[k], {k, 2^(n-1), 2^n-1}]; Flatten[ Table[ row[n], {n, 1, 5}]] (* Jean-François Alcover, Jan 24 2012 *)
scc[n_]:=Join@@Position[Reverse[IntegerDigits[n, 2]], 1];
Table[Reverse[scc[n]-Range[Length[scc[n]]]+1], {n, 0, 20}] (* Gus Wiseman, Jan 17 2023 *)
CROSSREFS
KEYWORD
tabf,nice,nonn
AUTHOR
Alford Arnold, Dec 10 2006
EXTENSIONS
Edited by Franklin T. Adams-Watters, Jun 11 2009
STATUS
approved